Certification Problem

Input (TPDB TRS_Standard/Strategy_removed_AG01/#4.28)

The rewrite relation of the following TRS is considered.

half(0)	→	0	(1)
half(s(0))	→	0	(2)
half(s(s(x)))	→	s(half(x))	(3)
bits(0)	→	0	(4)
bits(s(x))	→	s(bits(half(s(x))))	(5)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Constant to Unary

Every constant is turned into a unary function symbol to obtain the TRS

half(0'(x))	→	0'(x)	(6)
half(s(0'(x)))	→	0'(x)	(7)
half(s(s(x)))	→	s(half(x))	(3)
bits(0'(x))	→	0'(x)	(8)
bits(s(x))	→	s(bits(half(s(x))))	(5)

1.1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS

0'(half(x))	→	0'(x)	(9)
0'(s(half(x)))	→	0'(x)	(10)
s(s(half(x)))	→	half(s(x))	(11)
0'(bits(x))	→	0'(x)	(12)
s(bits(x))	→	s(half(bits(s(x))))	(13)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals

[0'(x₁)]	=	1 · x₁
[half(x₁)]	=	1 · x₁
[s(x₁)]	=	1 · x₁
[bits(x₁)]	=	1 · x₁ + 1

all of the following rules can be deleted.

0'(bits(x))

→

0'(x)

(12)

1.1.1.1 Switch to Innermost Termination

The TRS is overlay and locally confluent:

10

Hence, it suffices to show innermost termination in the following.

1.1.1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

0'^#(half(x))	→	0'^#(x)	(14)
0'^#(s(half(x)))	→	0'^#(x)	(15)
s^#(s(half(x)))	→	s^#(x)	(16)
s^#(bits(x))	→	s^#(half(bits(s(x))))	(17)
s^#(bits(x))	→	s^#(x)	(18)

1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair

s^#(bits(x)) → s^#(x) (18)

s^#(s(half(x))) → s^#(x) (16)

1.1.1.1.1.1.1 Usable Rules Processor

We restrict the rewrite rules to the following usable rules of the DP problem.

There are no rules.

1.1.1.1.1.1.1.1 Innermost Lhss Removal Processor

We restrict the innermost strategy to the following left hand sides.

s(s(half(x0)))

s(bits(x0))

1.1.1.1.1.1.1.1.1 Size-Change Termination

Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.

s^#(bits(x)) → s^#(x) (18)

1 > 1

s^#(s(half(x))) → s^#(x) (16)

1 > 1

As there is no critical graph in the transitive closure, there are no infinite chains.
The 2^nd component contains the pair

0'^#(s(half(x))) → 0'^#(x) (15)

0'^#(half(x)) → 0'^#(x) (14)

1.1.1.1.1.1.2 Usable Rules Processor

We restrict the rewrite rules to the following usable rules of the DP problem.

There are no rules.

1.1.1.1.1.1.2.1 Innermost Lhss Removal Processor

We restrict the innermost strategy to the following left hand sides.

s(s(half(x0)))

s(bits(x0))

1.1.1.1.1.1.2.1.1 Size-Change Termination

Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.

0'^#(s(half(x))) → 0'^#(x) (15)

1 > 1

0'^#(half(x)) → 0'^#(x) (14)

1 > 1

As there is no critical graph in the transitive closure, there are no infinite chains.